Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define LL long long
int n;
int powi(int a, int b, int m)
{
    int c = 1;
    for (; b; b >>= 1, a = 1ll * a * a % m)
        if (b & 1) c = 1ll * c * a % m;
    return c;
}
const int BAS = 233;
const int MOD1 = 1e9 + 7;
const int MOD2 = 1e9 + 9;
const int BIV1 = powi(BAS, MOD1 - 2, MOD1);
const int BIV2 = powi(BAS, MOD2 - 2, MOD2);
const int N = 500000;
char st[N];
pair <int, int> V[N], S[N];
map <pair <int, int>, int> C;
LL res;
int main()
{
    scanf("%d", &n);
    scanf("%s", st + 1);
    V[0] = make_pair(1, 1);
    for (int i = 1; i <= n; ++ i)
    {
        if (st[i] == '>')
            V[i] = make_pair(1ll * V[i - 1].first * BAS % MOD1, 1ll * V[i - 1].second * BAS % MOD2),
            S[i] = S[i - 1];
        else if (st[i] == '<')
            V[i] = make_pair(1ll * V[i - 1].first * BIV1 % MOD1, 1ll * V[i - 1].second * BIV2 % MOD2),
            S[i] = S[i - 1];
        else if (st[i] == '+')
            V[i] = V[i - 1],
            S[i] = make_pair((S[i - 1].first + V[i].first) % MOD1, (S[i - 1].second + V[i].second) % MOD2);
        else if (st[i] == '-')
            V[i] = V[i - 1],
            S[i] = make_pair((S[i - 1].first - V[i].first + MOD1) % MOD1, (S[i - 1].second - V[i].second + MOD2) % MOD2);
    }
    for (int i = n; ~i; -- i)
    {
        res += C[make_pair((1ll * S[n].first * V[i].first + S[i].first) % MOD1,
            (1ll * S[n].second * V[i].second + S[i].second) % MOD2)];
        C[S[i]] ++;
    }
    printf("%lld\n", res);
}